Question
In parallelogram ABCD, point E is located on side AB, and line segment DE is drawn perpendicular to AB. Given that AD measures 5 cm, AB is 12 cm, and AE is 3 cm, determine the area of the parallelogram.
Since DE is perpendicular to AB, AED forms a right angled triangle whose hypotenuse is AD. Using Pythagoras theorem in triangle AED, So, ED2 = AD2 – AE2 = 52 – 32 = 25 – 9 = 16 So, ED = √16 = 4 cm Area of a parallelogram = base × height = AB × ED = 12 × 4 = 48 cm2More Geometry Questions
- Question 1
- Find the area of triangle having sides 5 m, 6 m, and 9 m.
- The length of the each side of an equilateral triangle is 7√3 cm . The area of circumcircle, (cm 2 ) is
- Length of a chord in a circle of radius 'r' cm, is 16 cm and distance between chord and centre of the circle is 15 cm. Find the value of (5r - 19).
- In the given figure, PQR is a triangle and quadrilateral ABCD is inscribed in it. QD = 2 cm, QC = 5 cm, CR = 3 cm. BR = 4 cm. PB = 6 cm. PA = 5 cm and AD =...
- If G is the centroid and AD, BE, CF are three medians of ∆ABC with area 96 cm² , then the area of ∆EFG is?
- Find the ratio between the area of the in-circle and the circum-circle of a square?
- If two complementary angles are in the ratio of 7:2, find the greater angle.
- Two circles have radii of 18 cm and 32 cm, and they touch each other externally. What is the length of a common tangent between them?
- The 2nd approximation to a root of the equation x2-x-1-0 in the interval (1, 2) by Bisection method will be:
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
