# ML Aggarwal Solutions Class 9 Mathematics Solutions for Pythagoras Theorem Exercise 12 in Chapter 12 - Pythagoras Theorem

Question 9 Pythagoras Theorem Exercise 12

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. }

Taking square root on both sides

\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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