GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 07 Dec 2019, 10:37 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # The hypotenuse of a right triangle is 10 cm. What is the

Author Message
TAGS:

### Hide Tags

Manager  Joined: 28 Oct 2009
Posts: 60
The hypotenuse of a right triangle is 10 cm. What is the  [#permalink]

### Show Tags

19
180 00:00

Difficulty:   65% (hard)

Question Stats: 56% (01:42) correct 44% (01:38) wrong based on 2680 sessions

### HideShow timer Statistics

The hypotenuse of a right triangle is 10 cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length.

I know that (2) is sufficient but I am having difficulty with (1).
Math Expert V
Joined: 02 Sep 2009
Posts: 59588

### Show Tags

31
30
marcusaurelius wrote:
The hypotenuse of a right triangle is 10cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length.

I know that (2) is sufficient but I am having difficulty with (1).

For (1):
Let the legs be $$x$$ and $$y$$.

Given: $$h=10$$ and $$\frac{xy}{2}=25$$ (area of the right triangle $$area=\frac{leg_1*leg_2}{2}$$) --> $$h^2=100=x^2+y^2$$ (Pythagoras) and $$xy=50$$.

Q: $$P=x+y+10=?$$ So we have to calculate the value $$x+y$$.

Square $$x+y$$ --> $$(x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200$$ --> $$x+y=\sqrt{200}$$.

Thus $$P=x+y+10=\sqrt{200}+10$$.

Sufficient.
_________________
CEO  S
Joined: 20 Mar 2014
Posts: 2561
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44 GPA: 3.7
WE: Engineering (Aerospace and Defense)
The hypotenuse of a right triangle is 10 cm. What is the  [#permalink]

### Show Tags

3
2
luisnavarro wrote:

Hi Bunuel, I also have the same doubt, Knowing 2 of the 3 (Diagonal, perimeter, or area), we can find the third one por Triangles, Squares and rectangles? Is that correct. Could you help us?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700

Let me try to answer this question.

1. For ANY triangle, perimeter = P = a+b+c ---> so given 2 of the 3 sides and P, you can definitely calculate the 3rd side. For triangles that are not right angled or equilateral or isosceles, you would 1 angle or some sort of a relation between the sides and the area or a combination of these to sufficiently answer DS questions for these triangles.

i) For right triangles you also have the hypotenuse ^2 = base^2 + perpendicular ^2 as another equation. Thus for a right angled triangle, you have 2 equations.

ii) Similarly for isosceles and equilateral triangles (by drawing an a right angle, as shown in picture attached), you divide these 2 triangles into 2 same or congruent right angled triangles. So, for these 2 triangles as well you have 2 equations as mentioned above.

FYI, also remember that for right angled triangles and isosceles/equilateral triangles, you can calculate the 3rd side or the perimeter if the area is given as shown in the question discussed in this thread (by applying the relation, $$(a+b)^2 = a^2+b^2+2ab$$ and noticing that 0.5*ab represents the area of the triangle).

2. For rectangles and squares, the diagonals of these 2 shapes, divide these rectangles and squares into 2 congruent or same right angled triangles. The points mentioned above in 1) thus also apply these shapes as well.

Hope this answers your question. Let me know if there are any other questions.
Attachments Rect, Square and Triangle.jpg [ 36.37 KiB | Viewed 53824 times ]

##### General Discussion
Manager  Joined: 08 Jun 2011
Posts: 77

### Show Tags

1
1
Bunuel wrote:
marcusaurelius wrote:
The hypotenuse of a right triangle is 10cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length.

I know that (2) is sufficient but I am having difficulty with (1).

For (1):
Let the legs be $$x$$ and $$y$$.

Given: $$h=10$$ and $$\frac{xy}{2}=25$$ (area of the right triangle $$area=\frac{leg_1*leg_2}{2}$$) --> $$h^2=100=x^2+y^2$$ (Pythagoras) and $$xy=50$$.

Q: $$P=x+y+10=?$$ So we have to calculate the value $$x+y$$.

Square $$x+y$$ --> $$(x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200$$ --> $$x+y=\sqrt{200}$$.

Thus $$P=x+y+10=\sqrt{200}+10$$.

Sufficient.

Bunuel, if we factored out the equation in (A) and ended up with two different values for X, does that make (A) insufficient? I think it would. Would the GMAT be cruel enough to pull this trick on us? Because if we have to solve it all the way to determine the value of x+y then it's really a PS problem rather than a DS problem.

I am asking this because I was wondering whether we really needed to solve it for A. If we end up with an non linear equation in a similar DS statement, do we really have to solve it? Aren't we supposed to just determine if we have sufficient data to solve the problem.
Math Expert V
Joined: 02 Sep 2009
Posts: 59588

### Show Tags

6
4
Bunuel wrote:
marcusaurelius wrote:
The hypotenuse of a right triangle is 10cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length.

I know that (2) is sufficient but I am having difficulty with (1).

For (1):
Let the legs be $$x$$ and $$y$$.

Given: $$h=10$$ and $$\frac{xy}{2}=25$$ (area of the right triangle $$area=\frac{leg_1*leg_2}{2}$$) --> $$h^2=100=x^2+y^2$$ (Pythagoras) and $$xy=50$$.

Q: $$P=x+y+10=?$$ So we have to calculate the value $$x+y$$.

Square $$x+y$$ --> $$(x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200$$ --> $$x+y=\sqrt{200}$$.

Thus $$P=x+y+10=\sqrt{200}+10$$.

Sufficient.

Bunuel, if we factored out the equation in (A) and ended up with two different values for X, does that make (A) insufficient? I think it would. Would the GMAT be cruel enough to pull this trick on us? Because if we have to solve it all the way to determine the value of x+y then it's really a PS problem rather than a DS problem.

I am asking this because I was wondering whether we really needed to solve it for A. If we end up with an non linear equation in a similar DS statement, do we really have to solve it? Aren't we supposed to just determine if we have sufficient data to solve the problem.

First of all, we are asked to find the value of $$x+y$$ not $$x$$ or $$y$$ and that's what we did: $$x+y=\sqrt{200}$$. Now, $$(x+y)^2=200$$ has two solutions: $$-\sqrt{200}$$ and $$\sqrt{200}$$, but the first one is not valid since $$x$$ and $$y$$ must be positive. So there is only one acceptable numerical value of $$x+y$$ possible, regardles of the individual values of $$x$$ and $$y$$. Which makes this statement sufficient.

Next, even if we were asked to find the value of $$x$$ or $$y$$ then yes, $$xy=50$$ and $$x^2+y^2=100$$ gives two values for $$x$$ and $$y$$ BUT in this case the answer would still be sufficient since again one of the values would be negative, thus not a valid solution for a length.

As for the solving DS questions: when dealing with DS problems try to avoid calculations as much as possible. Remember DS problems do not ask you to solve, but rather to determine if you are ABLE to solve and in many cases you can determine that a statement is sufficient without working out all of the math. So if you are able to see from $$xy=50$$ and $$x^2+y^2=100$$ that it's possible to solve for $$x+y$$, then you don't need to actually do the math.

Hope it's clear.
_________________
Manager  Joined: 27 Oct 2011
Posts: 116
Location: United States
Concentration: Finance, Strategy
GPA: 3.7
WE: Account Management (Consumer Products)
Re: The hypotenuse of a right triangle is 10cm. What is the  [#permalink]

### Show Tags

stmt 1 we know that bh=50 and x^2 + y^2 = 100. Also, we need to solve for x+y+100. If we were to make (x+y)^2 and then solve out from there we can manipulate it so we can create an equation to what we know so far.
Manager  Joined: 18 May 2014
Posts: 54
Location: United States
Concentration: General Management, Other
GMAT Date: 07-31-2014
GPA: 3.99
WE: Analyst (Consulting)
Re: The hypotenuse of a right triangle is 10 cm. What is the  [#permalink]

### Show Tags

1
2
x^2+y^2= 100 (Pyth Theorem).....1)
2xy=100.........................2)
(x+y)^2= 200
or x+y=10 sqroot 2 (Hence statement 1 is sufficient)
or Perimeter = x+y+10 = 10(1+sq.rt2)
Statement 2. 2x^2=100; x=5 sqrt2
Perimeter = 10 (1+sqrt2)..(Hence, statement 2 is sufficient)
As both statements are sufficient. Hence, D
Intern  Joined: 17 May 2014
Posts: 36
Re: The hypotenuse of a right triangle is 10 cm. What is the  [#permalink]

### Show Tags

1
marcusaurelius wrote:
The hypotenuse of a right triangle is 10 cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length.

I know that (2) is sufficient but I am having difficulty with (1).

The best approach to tackle statement questions in DS is as follows:

Step 1: Convert all the alphabetical statements in algebraic statements
Step 2: Reduce the number of variable to minimum
Step 3: Check how many variables are left. You may probably need that many statements to solve the questions but you might need lesser number of statements to answer the question.

Caution: Don't waste your time in solving the question. You have to analyse the data sufficiency and not solve the question.

Lets see how this approach works for this question:

Step 1: A right triangle is given with hypotenuses equal 10. A triangle has three side. Let the other two sides be a and b

We know in a right triangle a^2 + b^2 =100

and we have to determine perimeter = a + b + 10

Step 2: We introduced 2 variables a and b above. Can we reduce the constraint equation to one variable. Yes, we can.

p = a + \sqrt{100-a^2} + 10

Thus, we are left with one variable.

Step 3: Let us look at the statements and see if we can decipher this one variable

1) The area of the triangle is 25 square centimeters.
Area of a right triangle is 1/2*a*b = 25
But we know the relationship between a and b as well. Therefore, we can calculate a and hence perimeter. Thus, this is a sufficient condition.\

2) The 2 legs of the triangle are of equal length.

This means a= b

Again, we can calculate a here and thus can find out the perimeter. Again sufficient.

Thus, the correct answer is D.

Note: We didn't solve for a or p nor did we enter any algebra here
Manager  Joined: 15 Aug 2013
Posts: 227

### Show Tags

1
Bunuel wrote:
marcusaurelius wrote:
The hypotenuse of a right triangle is 10cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length.

I know that (2) is sufficient but I am having difficulty with (1).

For (1):
Let the legs be $$x$$ and $$y$$.

Given: $$h=10$$ and $$\frac{xy}{2}=25$$ (area of the right triangle $$area=\frac{leg_1*leg_2}{2}$$) --> $$h^2=100=x^2+y^2$$ (Pythagoras) and $$xy=50$$.

Q: $$P=x+y+10=?$$ So we have to calculate the value $$x+y$$.

Square $$x+y$$ --> $$(x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200$$ --> $$x+y=\sqrt{200}$$.
Thus $$P=x+y+10=\sqrt{200}+10$$.

Sufficient.

Hi Bunuel,

Once you outlined the steps above, it's rather easy to solve. My question lies with the strategy -- how did you make the leap to "square" (x+y). What is the problem tipped you off that you had to solve a quadratic or at least, re-arrange it? Highlighted the area in question above.

Thanks!
Math Expert V
Joined: 02 Sep 2009
Posts: 59588

### Show Tags

1
6
russ9 wrote:
Bunuel wrote:
marcusaurelius wrote:
The hypotenuse of a right triangle is 10cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length.

I know that (2) is sufficient but I am having difficulty with (1).

For (1):
Let the legs be $$x$$ and $$y$$.

Given: $$h=10$$ and $$\frac{xy}{2}=25$$ (area of the right triangle $$area=\frac{leg_1*leg_2}{2}$$) --> $$h^2=100=x^2+y^2$$ (Pythagoras) and $$xy=50$$.

Q: $$P=x+y+10=?$$ So we have to calculate the value $$x+y$$.

Square $$x+y$$ --> $$(x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200$$ --> $$x+y=\sqrt{200}$$.
Thus $$P=x+y+10=\sqrt{200}+10$$.

Sufficient.

Hi Bunuel,

Once you outlined the steps above, it's rather easy to solve. My question lies with the strategy -- how did you make the leap to "square" (x+y). What is the problem tipped you off that you had to solve a quadratic or at least, re-arrange it? Highlighted the area in question above.

Thanks!

It should come with practice...

We know the values of xy and x^2+y^2, while need to get the value of x+y. Now, if you add twice xy to x^2+y^2 you get the square of x+y, hence squaring x+y is quite natural thing to do.

Similar questions to practice:
if-p-is-the-perimeter-of-rectangle-q-what-is-the-value-of-p-135832.html
if-the-diagonal-of-rectangle-z-is-d-and-the-perimeter-of-104205.html
what-is-the-area-of-rectangular-region-r-166186.html
what-is-the-perimeter-of-rectangle-r-96381.html

Hope this helps.
_________________
Manager  Joined: 15 Aug 2013
Posts: 227
Re: The hypotenuse of a right triangle is 10 cm. What is the  [#permalink]

### Show Tags

Bunuel wrote:
marcusaurelius wrote:
The hypotenuse of a right triangle is 10cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length.

I know that (2) is sufficient but I am having difficulty with (1).

For (1):
Let the legs be $$x$$ and $$y$$.

Given: $$h=10$$ and $$\frac{xy}{2}=25$$ (area of the right triangle $$area=\frac{leg_1*leg_2}{2}$$) --> $$h^2=100=x^2+y^2$$ (Pythagoras) and $$xy=50$$.

Q: $$P=x+y+10=?$$ So we have to calculate the value $$x+y$$.

Square $$x+y$$ --> $$(x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200$$ --> $$x+y=\sqrt{200}$$.

Thus $$P=x+y+10=\sqrt{200}+10$$.

Sufficient.

Hi Bunuel,

Please correct me if i'm wrong but it seems though if we know 2 of the 3 (Diagonal, perimeter, or area), we can find the third one. Is that correct?

Does this work for rectangles/squares and other figures as well?

Also, can you suggest similar problems -- one where the perimeter is given and we are required to find something else?

Thanks!
Manager  Joined: 07 Apr 2014
Posts: 98
The hypotenuse of a right triangle is 10 cm. What is the  [#permalink]

### Show Tags

marcusaurelius wrote:
The hypotenuse of a right triangle is 10 cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length.

I know that (2) is sufficient but I am having difficulty with (1).

i applied like this..

1:1:\sqrt{2} for option 1 .. since hypo is 10 then

other 2 are 5\sqrt{2}

(5\sqrt{2} *5\sqrt{2}) /2 = 25.. suff.

1:\sqrt{3}:2 wont suffice here ...

option b . same logic

1:1:\sqrt{2}

Hence D.
Intern  Joined: 11 Nov 2013
Posts: 2
Re: The hypotenuse of a right triangle is 10 cm. What is the  [#permalink]

### Show Tags

Also considering A.

We know that in a right triangle
[(SHORT SIDE#1)x(SHORT SIDE#2)]/ HYPOTENUSE = HEIGHT

from that we can compute the sum of the two short sides and so the perimeter.

Hope it helps!
Intern  Joined: 31 Jul 2014
Posts: 40
Concentration: Finance, Technology
Schools: Owen '17 (M$) Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink] ### Show Tags x and y are the two unknown sides... we can determine from the stem that x^2 + y^2 = 10^2 1.) we know triangle area = 1/2(base)(height) so 1/2(x)(y)=25. from this we can determine that (x)(y) = 50. x= 50/y. using equation above we have 2 equations 2 unknowns. sufficient. 2.) x^2 + x^2 = 10^2. sufficient. Manager  G Joined: 13 Oct 2013 Posts: 132 Concentration: Strategy, Entrepreneurship The hypotenuse of a right triangle is 10 cm. What is the [#permalink] ### Show Tags 1 This is correct, i used the triangle 45:45:90 and calculated the sides 1:1:root 2 each side =5 root 2 perimeter=10 root 2+10 your answer is correct . Intern  Joined: 23 Jun 2015 Posts: 11 Schools: YLP '16 GPA: 4 Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink] ### Show Tags 2 Bunuel wrote: marcusaurelius wrote: The hypotenuse of a right triangle is 10cm. What is the perimeter, in centimeters, of the triangle? (1) The area of the triangle is 25 square centimeters. (2) The 2 legs of the triangle are of equal length. I know that (2) is sufficient but I am having difficulty with (1). For (1): Let the legs be $$x$$ and $$y$$. Given: $$h=10$$ and $$\frac{xy}{2}=25$$ (area of the right triangle $$area=\frac{leg_1*leg_2}{2}$$) --> $$h^2=100=x^2+y^2$$ (Pythagoras) and $$xy=50$$. Q: $$P=x+y+10=?$$ So we have to calculate the value $$x+y$$. Square $$x+y$$ --> $$(x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200$$ --> $$x+y=\sqrt{200}$$. Thus $$P=x+y+10=\sqrt{200}+10$$. Sufficient. How: 1) is sufficient.. But x and y are coming out to be imaginary, so such sides of a right triangle don't exist.....so not point of calculating perimeter from (1) thought we can calculate ...x+y....because imaginary parts cancelled out in addition...but if x and y are not real how can we think of perimeter... Math Expert V Joined: 02 Sep 2009 Posts: 59588 The hypotenuse of a right triangle is 10 cm. What is the [#permalink] ### Show Tags himanshutyagi99 wrote: Bunuel wrote: marcusaurelius wrote: The hypotenuse of a right triangle is 10cm. What is the perimeter, in centimeters, of the triangle? (1) The area of the triangle is 25 square centimeters. (2) The 2 legs of the triangle are of equal length. I know that (2) is sufficient but I am having difficulty with (1). For (1): Let the legs be $$x$$ and $$y$$. Given: $$h=10$$ and $$\frac{xy}{2}=25$$ (area of the right triangle $$area=\frac{leg_1*leg_2}{2}$$) --> $$h^2=100=x^2+y^2$$ (Pythagoras) and $$xy=50$$. Q: $$P=x+y+10=?$$ So we have to calculate the value $$x+y$$. Square $$x+y$$ --> $$(x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200$$ --> $$x+y=\sqrt{200}$$. Thus $$P=x+y+10=\sqrt{200}+10$$. Sufficient. How: 1) is sufficient.. But x and y are coming out to be imaginary, so such sides of a right triangle don't exist.....so not point of calculating perimeter from (1) thought we can calculate ...x+y....because imaginary parts cancelled out in addition...but if x and y are not real how can we think of perimeter... $$x+y=\sqrt{200}$$ so it's an irrational number, not imaginary number. _________________ Senior Manager  B Joined: 10 Mar 2013 Posts: 461 Location: Germany Concentration: Finance, Entrepreneurship Schools: WHU MBA"20 (A$)
GMAT 1: 580 Q46 V24 GPA: 3.88
WE: Information Technology (Consulting)
The hypotenuse of a right triangle is 10 cm. What is the  [#permalink]

### Show Tags

1
We can solve it using topic about similar triangles
1)Area= (10*Height)/2 = 25 --> Height=5, so if this 3 triangles are similar, then x/5=5/10-x, x^2-10x+25=0 --> x=5
The hypotenuses of small traingles => 1st: x^2+25, 2nd:125-20x+x^2..... So we can calculate Perimiter after pluging X=5 in the stated equations (no need to calculate) SUFFICIENT

2) 2 Legs are equal => pythagoras theorem: x^2+x^2=100, x=5*sqrt 2, so again it's enougj to calculate the perimeter (10+10*sqrt 2)
Attachments 128.jpg [ 22.05 KiB | Viewed 53860 times ]

Intern  Joined: 24 Jun 2015
Posts: 45
Re: The hypotenuse of a right triangle is 10 cm. What is the  [#permalink]

### Show Tags

russ9 wrote:
Bunuel wrote:
marcusaurelius wrote:
The hypotenuse of a right triangle is 10cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length.

I know that (2) is sufficient but I am having difficulty with (1).

For (1):
Let the legs be $$x$$ and $$y$$.

Given: $$h=10$$ and $$\frac{xy}{2}=25$$ (area of the right triangle $$area=\frac{leg_1*leg_2}{2}$$) --> $$h^2=100=x^2+y^2$$ (Pythagoras) and $$xy=50$$.

Q: $$P=x+y+10=?$$ So we have to calculate the value $$x+y$$.

Square $$x+y$$ --> $$(x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200$$ --> $$x+y=\sqrt{200}$$.

Thus $$P=x+y+10=\sqrt{200}+10$$.

Sufficient.

Hi Bunuel,

Please correct me if i'm wrong but it seems though if we know 2 of the 3 (Diagonal, perimeter, or area), we can find the third one. Is that correct?

Does this work for rectangles/squares and other figures as well?

Also, can you suggest similar problems -- one where the perimeter is given and we are required to find something else?

Thanks!

Hi Bunuel, I also have the same doubt, Knowing 2 of the 3 (Diagonal, perimeter, or area), we can find the third one por Triangles, Squares and rectangles? Is that correct. Could you help us?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700
Intern  Joined: 24 Jun 2015
Posts: 45
Re: The hypotenuse of a right triangle is 10 cm. What is the  [#permalink]

### Show Tags

Engr2012 wrote:
luisnavarro wrote:

Hi Bunuel, I also have the same doubt, Knowing 2 of the 3 (Diagonal, perimeter, or area), we can find the third one por Triangles, Squares and rectangles? Is that correct. Could you help us?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700

Let me try to answer this question.

1. For ANY triangle, perimeter = P = a+b+c ---> so given 2 of the 3 sides and P, you can definitely calculate the 3rd side. For triangles that are not right angled or equilateral or isosceles, you would 1 angle or some sort of a relation between the sides and the area or a combination of these to sufficiently answer DS questions for these triangles.

i) For right triangles you also have the hypotenuse ^2 = base^2 + perpendicular ^2 as another equation. Thus for a right angled triangle, you have 2 equations.

ii) Similarly for isosceles and equilateral triangles (by drawing an a right angle, as shown in picture attached), you divide these 2 triangles into 2 same or congruent right angled triangles. So, for these 2 triangles as well you have 2 equations as mentioned above.

FYI, also remember that for right angled triangles and isosceles/equilateral triangles, you can calculate the 3rd side or the perimeter if the area is given as shown in the question discussed in this thread (by applying the relation, $$(a+b)^2 = a^2+b^2+2ab$$ and noticing that 0.5*ab represents the area of the triangle).

2. For rectangles and squares, the diagonals of these 2 shapes, divide these rectangles and squares into 2 congruent or same right angled triangles. The points mentioned above in 1) thus also apply these shapes as well.

Hope this answers your question. Let me know if there are any other questions.

Thanks a lot Engr2012.

That mean that this kind of question (Given area and diagonal of the examples above mentioned) in DS are "automatic", for instance; No matter system equation is quadratic always is going to be one answer and would be sufficient (I do not need to check the specific roots of the quadratic). Is it correct?

Regards.

Luis Navarro
Looking for 700 Re: The hypotenuse of a right triangle is 10 cm. What is the   [#permalink] 18 Jul 2015, 16:44

Go to page    1   2   3    Next  [ 47 posts ]

Display posts from previous: Sort by

# The hypotenuse of a right triangle is 10 cm. What is the  