Question

Q.4. The parallel sides of a trapezium are
23 cm and 11 cm long, while its non-
parallel sides are 10 cm each. Find the area
of the trapezium.​

Answer 1

$\large\bf\underline{Given:-}$

• The parellel side's of trapezium are 23 cm and 11cm.
• Non - parellel side's are 10cm each

$\large\bf\underline {\:To \: find:-}$

• Area of trapezium

$\huge\rm\underline{\bold\:Solution:-}$

# Diagram was in the attachment.

Let :-

• PQ = 23 cm
• SR = 11cm
• PS = QS = 10cm

Now,

≫ AQ = PQ - PA

≫ AQ = (23 - 11)cm

≫ AQ = 12cm

PB ⊥ PQ, So B is the mid point of AQ.

• AB = ½ × AQ
• AB = ½ × 12
• AB = 6 cm

In right angled triangle ∆ RAB,

• AR = 10cm
• AB = 6cm

By Pythagoras theorem:-

➝ RB = √AR² - AB²

➝ RB = √10² - 6²

➝ RB = √100-36

➝ RB = √64

➝ RB = 8cm

So, the distance between the parellel sides of trapezium is 8cm.

✝️Area of trapezium:-

= ½(sum of || sides)×(distance between the parellel sides)

»» ½(23 + 11) × 8

»» ½ × 34 × 8

»» 17 × 8

»» 136cm²

Hence ,

✝️ Area of trapezium = 136cm².

Answer 2

Answer:-

$\sf{The \ area \ of \ trapezium \ is \ 136 \ cm^{2}}$

Given:

1. For trapezium,

• Parallel sides are 23 cm and 11 cm long.

• Non parallel sides are 10 cm each.

To find:

• Area of the trapezium.

Solution:

$\sf{\underline{\underline{Construction:}}}$

$\sf{1. \ Draw \ AE \ parallel \ to \ BC.}$

$\sf{2. \ Draw \ AF \ perpendicular \ to \ CD.}$

$\sf{In \ quadrilateral \ ABCD,}$

$\sf{AB=11 \ cm,}$

$\sf{BC=10 \ cm,}$

$\sf{CD=23 \ cm,}$

$\sf{EC=11 \ cm,}$

$\sf{AE=10 \ cm,}$

$\sf{...\because{Distance \ between \ two \ parallel \ lines}}$

$\sf{is \ constant. }$

$\sf{DE=DC-EC=23-11...[D-E-C]}$

$\sf{\therefore{DE=12 \ cm}}$

$\sf{In \ \triangle \ AED,}$

$\sf{DF=\frac{1}{2}\times \ DE}$

$\sf{DF=\frac{1}{2}\times12=6 \ cm}$

$\sf{...[\triangle \ AED \ is \ isosceles \ triangle.]}$

$\sf{By \ Apollonius \ theorem,}$

$\sf{AD^{2}+AE^{2}=2AF^{2}+2DF^{2}}$

$\sf{\therefore{10^{2}+10^{2}=2(AF^{2}+6^{2})}}$

$\sf{\therefore{2\times10^{2}=2(AF^{2}+6^{2})}}$

$\sf{\therefore{10^{2}=AF^{2}+6^{2}}}$

$\sf{\therefore{AF^{2}=100-36}}$

$\sf{\therefore{AF^{2}=64}}$

$\sf{On \ taking \ square \ root \ of \ both \ sides. }$

$\sf{AF=8 \ cm}$

$\sf{But, \ AF \ is \ perpendicular \ to \ DC}$

$\sf{\therefore{Height=AF=8 \ cm}}$

$\boxed{\sf{Area \ of \ trapezium=\frac{1}{2}\times(Sum \ of \ parallel \ sides)\times \ Height}}$

$\sf{\therefore{Area \ of \ trapezium=\frac{1}{2}\times(11+23)\times8}}$

$\sf{\therefore{Area \ of \ trapezium=4\times34}}$

$\sf{\therefore{Area \ of \ trapezium=136 \ cm^{2}}}$

$\sf\purple{\tt{\therefore{The \ area \ of \ trapezium \ is \ 136 \ cm^{2}}}}$