In the given figure, if DE || BC, area of △ADE = 1 and area of △ADC =

Question

In the given figure, if DE || BC, area of △ADE = 1 and area of △ADC = 4, then find area of △DBC. Capture.PNG A. 1 B. 4 C. 6 D. 12 E. 16

chetan2u · Answer

Height of each of △ADE, △ADC, and △CDE is same, say h.

area of △ADE =  \frac{1}{2}*h*AE=  1 and 
area of △ADC =  \frac{1}{2}*h*CE=  4

So AE:CE=1:4

As △ADE and △ABC are similar,  the area of the two triangles will be square of the ratio of sides.  
So area of △ADE : area of △ABC =  1^2:4^2=1:16

area of △DBC = area of △ABC - area of △ADC = 16-4 = 12

D

You can also estimate that the area of DEC has to be at least 2-3 times that of ADC. Hence, only D and E left.

Harshjha001 · Answer

How can we confirm if DE is the height ?

v12345 · Answer

Area of DBC is 12.
Hence, OA is (D).

Fdambro294 · Answer

Obviously, chetan2u has the quickest and easiest way to get to the answer. But an alternative method:

Formula: one way to find the Area of a triangle is to multiply two adjacent sides —- then take that product and multiply it by the (SINE of the included angle between those sides) such that:

Area of Triangle = (1/2) (a) (b) (SIN of included angle)

Where A and B are the adjacent sides and the included angle is the angle found BETWEEN those 2 sides

(Step 1) focusing on triangle ADE and triangle ADC

We are given the actual Areas of these 2 triangles.

Since AE and AC are both on the same line, we can use those sides as the respective Bases for each triangle.

Since there exists only one perpendicular height from a single point to a line, the Perpendicular Height drawn from vertex D will be the same for both triangles.

Using the regular Area of a Triangle formula:

Area ADE = (1/2) (h) (AE) = 1

Area ADC = (1/2) (h) (AC) = 4

Dividing the 2 equations we get the RATIO of the 2 Bases:

(AE) : (AC) = 1 : 4

(Step 2). We are given that DE is parallel to BC

This makes triangle ADE a Similar Triangle to the whole triangle ABC

The corresponding sides will be in a Constant Ratio ——-> which we found in step 1

Corresponding sides of the similar triangles ——-> AE : AC = 1 : 4

The other corresponding sides will be in the same ratio and we can use K to stand for the unknown ratio multiplier

DE : BC = 1k : 4k

AD : AB = 1k : 4k

Side DB (one of the sides of triangle DBC, the area we need to find) will be equal to the difference between Side AB and Side AD ——-> 4k - 1k = 3k

Thus, in summary:

Triangle ADE:

AD = 1k

DE = 1k

Included Angle between these sides ———Angle < D

Triangle DBC:

DB = 3k

BC = 4k

Included Angle between these sides ——— Angle < B

(Step 3) we can use the formula listed in the beginning to find the Area of triangle ADE and then finally the target Triangle, triangle DBC

Area of Triangle ADE = (1/2) (AD) (DE) (SIN of included angle D) = 1

Or

(1/2) (1k) (1k) (Sin X) = 1

(k^2) (Sin X) = 2

we are told that DE is parallel to BC ——-> making Angle < D and Angle < B corresponding CONGRUENT angles in the Similar triangles ADE and ABC

SIN of Angle < D = SIN of Angle < B = (Sin X)

**Area of triangle DBC =

(1/2) (DB) (BC) (Sin of Angle < B) =

(1/2) (3k) (4k) (Sin X) = ?

Or

(6) (k^2) (Sin X) = ?

From the above Area of triangle ADE, we found:

(k^2) (Sin X) = 2

Substituting this in for our Area of triangle DBC we get:

(6) (2) = 12

Area of triangle DBC = 12

D

Posted from my mobile device

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In the given figure, if DE || BC, area of △ADE = 1 and area of △ADC =

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In the given figure, if DE || BC, area of △ADE = 1 and area of △ADC =

Prep Toolkit

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