Answer:
36cm²
Explanation:
From the sketched diagram of the Quadrilateral above, we have two triangles that are joined together, ∆PQR and ∆PSR
Both triangles are right angled triangles.
Step 1
We solve for ∆PQR
When solving for the sides of a right angled triangle, if two sides are given, we solve for the third sides using Pythagoras Theorem
Pythagoras Theorem =
c² = a² + b²
Where c = Longest side
For ∆PQR,
/PQ/ = 3 cm = a
/QR/ = 4 cm = b
/PR/ = c ??
c² = 3² + 4²
c² = 9 + 16
c² = 25
c = √25
c = 5cm
/PS/ = 5cm
We find the Area of this triangle using Heron's formula
= √s(s - a)(s - b)(s - c)
Where s = a + b + c/2
s = 3 + 4 + 5/2
= 12/2
s = 6
Area of the ∆PQR = √6(6 - 3)(6 - 4)(6 - 5)
= √6 × 3 × 2 × 1
= √36
= 6cm²
Area of the ∆PQR
Step 2
We solve for ∆PSR. Just like in Step 1, we solve for third side not given using Pythagoras Theorem
/PS/ = 13cm = c
/PR/ = 5cm = a
/SR/ = ?? = b
c² = a² + b²
13² = 5² + b²
b² = 13² - 5²
b² = 169 - 25
b² = 144
b = √144
b = 12cm
/SR/ = 12cm
We find the Area of this triangle using Heron's formula
= √s(s - a)(s - b)(s - c)
Where s = a + b + c/2
s = 5 + 12 + 13/2
= 30/2
s = 15
Area of the ∆PSR = √15(15 - 5)(15 - 12)(15 - 13)
= √15 × 10 × 3 × 2
= √900
= 30cm²
Area of the ∆PSR = 30 cm²
Step 3
We find the Area of the Quadrilateral
Area of the Quadrilateral (PQRS)= Area of ∆PQR + Area of ∆PSR
= 6cm² + 30cm²
= 36cm²
Therefore the Area of the Quadrilateral PQRS = 36cm²