an isosceles trapezium is such that its parallel sides are of length 24 cm and 48 cm respectively. whereas the non parallel sides are each 37 cm long. find the area of the trapezium [ At the back the answer is given 1260cm^2]
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Answer:
Step-by-step explanation:
Given ,
An isosceles trapezium such that its parallel sides are of length 24 cm and 48 cm respectively. Whereas non parallel sides are of 37 cm.
We have to find ,
Area of trapezium.
Solution :
Let us draw an isosceles trapezium ( isosceles trapezium means its non parallel sides are of equal length ) PQRS.
#Formula for finding area of isosceles trapezium :
[tex] \: \: \: \: \: \: \boxed{ \sf{ \dfrac{1}{2} \times ( sum \: of \: parallel \: side) \times height}}[/tex]
Here ,
In question , we are not provided the height of the trapezium. So for finding the height of trapezium, We are drawing,
Now ,
→ PQ = 48 cm
→ RS = 24 cm
→ PS = QR = 37 cm
Also ,
#Finding value of PA :
Now , finding height of trapezium using Pythagoras Theorem :
In Right angled triangle SAP :
→ (PS)² = (PA)² + (SA)²
→ (37)² = (12)² + (SA)²
→ (SA)² = (37)² - (12)²
For easy calculation , We are using an identity that is:
→ (SA)² = (37 - 12)(37 + 12)
→ (SA)² = 25 × 49
→ SA = √( 25 × 49 )
{ 25 is square root of 5 and 49 is square root of 7 }
→ SA = 5 × 7
→ SA = 35 cm
As we know the height of trapezium, let us substitute all the known values in the formula :
[tex] \: \: \: \: \: \sf{ \longmapsto \: \: \: \: \: \: \: \dfrac{1}{ \cancel{2}} \times ( \cancel{72)}\times 35 }[/tex]
[tex] \: \: \: \: \: \sf{ \longmapsto \: \: \: \: \: \: \: 36\times 35 }[/tex]
[tex] \: \: \: \: \: \sf{ \longmapsto \: \: \: \: \: \: \: \underline{ \boxed{ \bold{1260 \: cm {}^{2} }}}} \: \: \: \: \bigstar[/tex]
>>> Therefore, Area of given isosceles trapezium is "1260" cm².