find the area of shaded portion
Answers
Answer:
9 cm²
Step-by-step explanation:
∠A = 90°
In ΔAPQ, By Pythagoras theorem,
PQ² = AP² + AQ² = 3² + 3² = 18
PQ = 3√2 cm.
In ΔABC,
BC² = AB² + AC² = 6² + 6² = 72
BC = 6√2 cm.
we know that
BR+ SC = BC - RS = 6√2 - 3√2 = 3√2.
Considering triangles PBR and QCS, PB = QC and PR = QS thus BR = SC
=> 2BR = 3√2.
=> BR = 3/√2 = 3√2/2 cm = SC.
Now considering right angled triangle QCS,
using Pythagoras theorem.
QC² = QS² + SC².
3² = QS² + (3√2/2)²
QS² = 9 - 9/2 = 9/2 cm.
QS = 3/√2 cm = 3√2/2 cm.
Area of Shaded of region = Area of ΔABC - Area of Rectangle PQRS.
But to find Area of ΔABC we will need to find height of the triangle.
Draw line from vertex A to midpoint of BC meeting at point F.
=> BF = BC/2 = 6√2/2 = 3√2 cm.
Now ΔABF is a right angled triangle, by Pythagoras theorem,
AF² + BF² = AB²
AF² = 6² - (3√2)² = 36 - 18 = 18
AF = 3√2 cm.
Area of Isosceles triangle ABC = 1/2 * 3√2 * 6√2 = 18 cm².
Area of Rectangle PQRS = 3√2/2 * 3√2 = 9 cm².
So area of Shaded region = 18 - 9 = 9 cm².
