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ML Aggarwal Solutions Class 9 Mathematics Solutions for Pythagoras Theorem Exercise 12 in Chapter 12 - Pythagoras Theorem
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a) In figure (i) given below, AB || DC, BC = AD = 13 cm. AB = 22 cm and DC = 12cm. Calculate the
height of the trapezium ABCD.
(b) In figure (ii) given below, AB || DC, ∠ A = 90°, DC = 7 cm, AB = 17 cm and AC = 25 cm. Calculate BC.
(c) In figure (iii) given below, ABCD is a square of side 7 cm. if
AE = FC = CG = HA = 3 cm,
(i) prove that EFGH is a rectangle.
(ii) find the area and perimeter of EFGH.
(i) Given AB || DC, BC = AD = 13 cm.
AB = 22 cm and DC = 12cm
Here DC = 12
MN = 12 cm
AM = BN
AB = AM+MN+BN
22 = AM+12+AM [∵AM = BN]
2AM = 22-12 = 10
\begin{aligned} &\therefore \mathrm{AM}=10 / 2\\ &\therefore \mathrm{AM}=5 \mathrm{cm}\\ &\triangle \mathrm{AMD} \text { is a right triangle. }\\ &\mathrm{AD}^{2}=\mathrm{AM}^{2}+\mathrm{DM}^{2} \quad \text { [Pythagoras theorem] }\\ &13^{2}=5^{2}+\mathrm{DM}^{2}\\ &\therefore \mathrm{DM}^{2}=13^{2}-5^{2}\\ &\therefore \mathrm{DM}^{2}=169-25\\ &\therefore \mathrm{DM}^{2}=144 \end{aligned}
Taking square root on both sides,
DM = 12 cm
Hence the height of the trapezium is 12 cm.
(b) Given AB || DC, A = 90°, DC = 7 cm,
AB = 17 cm and AC = 25 cm
\triangle \mathrm{ADC} \text { is a right triangle. }
\begin{aligned} &\therefore \mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{DC}^{2} \quad[\text { Pythagoras theorem }]\\ &25^{2}=\mathrm{AD}^{2}+7^{2}\\ &\therefore \mathrm{AD}^{2}=25^{2}-7^{2}\\ &\therefore \mathrm{AD}^{2}=625-49\\ &\therefore \mathrm{AD}^{2}=576 \end{aligned}
Taking square root on both sides
AD = 24 cm
\begin{aligned} &\therefore \mathrm{CM}=24 \mathrm{cm} \quad[\because \mathrm{AB} \| \mathrm{CD}]\\ &\mathrm{DC}=7 \mathrm{cm}\\ &\therefore \mathrm{AM}=7 \mathrm{cm}\\ &\mathrm{BM}=\mathrm{AB}-\mathrm{AM}\\ &\therefore \mathrm{BM}=17-7=10 \mathrm{cm}\\ &\triangle \mathrm{BMC} \text { is a right triangle. }\\ &\therefore \mathrm{BC}^{2}=\mathrm{BM}^{2}+\mathrm{CM}^{2}\\ &\mathrm{BC}^{2}=10^{2}+24^{2}\\ &\mathrm{BC}^{2}=100+576\\ &\mathrm{BC}^{2}=676 \end{aligned}
Taking square root on both sides
BC = 26 cm
Hence the length of BC is 26 cm.
(c) (i)Proof:
Given ABCD is a square of side 7 cm.
So AB = BC = CD = AD = 7 cm
Also given AE = FC = CG = HA = 3 cm
BE = AB-AE = 7-3 = 4 cm
BF = BC-FC = 7-3 = 4 cm
GD = CD-CG = 7-3 = 4cm
DH = AD-HA = 7-3 = 4 cm
\begin{aligned} &\angle \mathrm{A}=90^{\circ}\\ &\triangle \mathrm{AHE} \text { is a right triangle. }\\ &\therefore \mathrm{HE}^{2}=\mathrm{AE}^{2}+\mathrm{AH}^{2} \quad[\text { Pythagoras theorem }]\\ &\therefore \mathrm{HE}^{2}=3^{2}+3^{2}\\ &\therefore \mathrm{HE}^{2}=9+9=18 \end{aligned}
HE = √(9×2) = 3√2 cm
Similarly GF = 3√2 cm
\begin{aligned} &\triangle \mathrm{EBF} \text { is a right triangle. }\\ &\therefore \mathrm{EF}^{2}=\mathrm{BE}^{2}+\mathrm{BF}^{2} \quad[\text { Pythagoras theorem] }\\ &\therefore \mathrm{EF}^{2}=4^{2}+4^{2}\\ &\therefore \mathrm{EF}^{2}=16+16=32 \end{aligned}
EF = √(16×2) = 4√2 cm
Similarly HG = 4√2 cm
Now join EG
\begin{aligned} &\operatorname{In} \triangle \mathrm{EFG}\\ &\mathrm{EG}^{2}=\mathrm{EF}^{2}+\mathrm{GF}^{2}\\ &\mathrm{EG}^{2}=(4 \sqrt{2})^{2}+(3 \sqrt{2})^{2}\\ &\mathrm{EG}^{2}=32+18=50 \end{aligned}
\begin{aligned} &\therefore \mathrm{EG}=\sqrt{50}=5 \sqrt{2} \mathrm{cm} \quad \ldots(\mathrm{i})\\ &\text { Join HF. }\\ &\text { Also } \mathrm{HF}^{2}=\mathrm{EH}^{2}+\mathrm{HG}^{2}\\ &=(3 \sqrt{2})^{2}+(4 \sqrt{2})^{2}\\ &=18+32=50 \end{aligned}
From (i) and (ii)
EG = HF
Diagonals of the quadrilateral are congruent. So EFGH is a rectangle.
Hence proved.
(ii)Area of rectangle EFGH = length × breadth
= HE ×EF
= 3√2×4√2
\begin{aligned} &=24 \mathrm{cm}^{2}\\ &\text { Perimeter of rectangle } \mathrm{EFGH}=2 \text { (length+breadth) } \end{aligned}
= 2×(4√2+3√2)
= 2×7√2
= 14√2 cm
Hence area of the rectangle is 24 \mathrm{cm}^{2} \text { and perimeter is } 14 \sqrt{2} \mathrm{cm}
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