Math, asked by JAGRITVOHRA2006, 10 months ago

ABC IS A RIGHT ANGLED TRIANGLE RIGHT ANGLED AT B BC=BA.D IS A POINT ON AC PRODUCED AND A LINE DEF CUTS CB AT E,AB AT F. IF ANGLE D IS 13 DEGREE AND ANGLE FAE =29 DEGREE THEN THE MEASURE OF ANGLE FEA IS ?

Answers

Answered by zeek861
4

Answer:

we are given

A diagram consists of three triangles ABC, ACD, and ADE – all of them are individually isosceles right triangles.

The area of ΔABC is 6.

To Find

We need to determine

The area of ΔADE.

Approach & Working

Let us assume the length of AB = n

As ΔABC is an isosceles right angle triangle, AB = BC = n

Therefore, AC = √(n2 + n2) = √2n

We also know that ΔACD is an isosceles right angle triangle.

Hence, AC = CD = √2n

Therefore, AD = √(2n2 + 2n2) = 2n

Finally, ΔADE is an isosceles right-angle triangle.

Hence, AD = DE = 2n

Thus, the area of ΔADE = ½ * 2n * 2n = 2n2

We are also given that area of ΔABC is 6.

Answered by somya2563
81

Step-by-step explanation:

Question:-

BC is a right triangle, right angled at B. BC = BA. D is a point on AC produced and a line DEF cuts CB at E, AB at F. IF ZD = 13º and ZFAE = 29, then

Solution :-

We are given,

A diagram consists of three triangles ABC ,ACD & ADC - all of them are individually as I isosceles right triangles.

The area of ∆ ABC = 6

You need to determine

The area of ∆ ADE

Approach and working

Let us assume the length of AB = n

As ABC is an isosceles right angle triangle ,AB = BC = n

Therefore, AC = √(n2+2n2) = 2n

we also know that ∆ADE is is an isosceles right angle triangle

hence, AD = DE = 2n

Thus, the area of ∆ADE = 1/2 × 2n × 2n = 2n2

We are also given that area of ∆ABC = 6

.

Hope it helpful .☺

@Somya2563

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