Question
The altitude drawn to the base of an isosceles triangle is 6 cm and the perimeter of the triangle is 36
The altitude drawn to the base of an isosceles triangle is 6 cm and the perimeter of the triangle is 36
cm. The area (in cm²) of the triangle is
More Geometry Questions
- AB and AC are the two tangents to a circle whose radius is 6 cm. If ∠ BAC = 60o, then what is the value (in cm) of √(AB2 + AC2)?
- Question 2
- The perimeter of a rectangle is 100 m, and its diagonal is 41 m. Find the area of the rectangle.
- Consider two concentric circles having radii 17 cm and 15 cm. What is the length (in cm) of the chord, of the bigger circle, which is a tangent to the smal...
- Given are three points A(2, 7), B(4, -1), C(- 2, 6) on a plane. What kind of triangle is formed by joining the points A, B, and C?
- Two chords, AB and CD, of a circle intersect at an external point E. Given that the lengths of BE and DE are 12 cm and 8 cm, respectively, and the total le...
- Two angles are complementary. The larger angle is 6° less than thrice the measure of the smaller angle. What is the measure of the larger angle?
- If O is the orthocentre of ΔABC , if ∠ BOC = 1250 then what is the measure of ∠ BAC?
- Find the area of triangle having sides 20 m, 21 m, and 29 m.
- Right triangle with sides 15 cm, 20 cm, 25 cm. Find radius of incircle.
Relevant for Exams:
Hey! Ask a query
Please enter email id
The email must be a valid email address.
Please enter Mobile Number
Please enter valid Mobile Number
Please enter your Doubt
Think You're Ready for RBI Grade B?
RBI Grade B 2026 Phase 1 Memory Based Paper
- 200 Questions with Detailed Solutions
- Section-wise Coverage (GA, English, Quant & Reasoning)
Let AB = AC = a cm. BD = DC = b cm. Altitude of isosceles triangle is also median. In right ∆ADC, 6² = a² - b² 36 = a² - b² ………… (i) Perimeter = 36 a + a + 2b = 36 2a + 2b = 36 a + b = 18 ………. (ii) On dividing (i) & (ii) we get, (a² - b² )/(a+b) = 36/18 = 2 ((a+b) (a-b))/((a+b)) = 2 a – b = 2 a + b = 18 on solving, 2a = 20 a = 10 b = 8 BC = 16 Area of ∆ABC = 1/2 × AD × BC = 1/2 × 6 × 16 = 48 cm²