Question

please solve this question fastly ​

Answer 1

Answer:

24

Step-by-step explanation:

please find attached

Answer 2

Let the sides of triangle be a, b and c.

c is the hypotenuse of the triangle.

Perimeter=120 units

Perimeter= sum of all sides

120=a+b+c

120=70+c [Sum of two shorter sides= a+b=70]

c=50

Area of triangle = 1/2 ×base ×altitude(h)

$\: \: \: \: \: \: \: \: \: \: = \frac{1}{2} \times a \times b$

= ab/2 ........(i)

In triangle PQR, right angled at Q, we have

PR²=PQ²+QR² => c²= a²+b²

c²= (a+b)²-2ab

(50)²=(70)²-2ab

2500=4900-2ab

2ab=4900-2500

2ab=2400

ab=1200

Substituting the value of the ab in (i)

Area of triangle = 1200/2=600 sq. units

Now

Area of triangle= 1/2 ×base ×altitude(h)

$600 = \frac{1}{2} \times 50 \times h \\ h = \frac{600}{25} \\ h = 24 \: units$