∆ ABC is isosceles with AB = AC = 7.5 cm, and BC = 9 cm (in Figure). The height AD from A to BC, is 6 cm. Find the area of ∆ ABC. What will be the height from C to AB i.e., CE?
∆ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (Fig 11.26
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex
15. If ABC is an isosceles triangle with AB = AC = 7.5cm and BC
13 cm APQR is isosceles with PQ=PR = 7.5 cm and QR=9 cm. The
ABC is an isosceles triangle in which ab is equal to AC 7.2 and BC
Ex 9.1, 8 - ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm
A triangle ABC has side of lengths a=14cm b=20cm c=7.5cm. What is
8. frac{1}{D} frac{13 mathrm{cm}}{mathrm{D} 13 mathrm{cm}} mathrm
NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex
ΔABC is isosceles with AB=AC=7cm and BC=9cm. The height AD from A
In a triangle ABC, ∠BAC = 90° and AD is perpendicular to BC. If
DeltaABC is isosceles with AB=AC=7.5cm and BC=9cm. The height AD
Δ ABC is isosceles with AB=AC=7.5 cm and BC = 9 cm Figure. The
