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**Short Answer Type Questions****Q1.If one diagonal of a trapezium divides the other diagonal in the ratio 1 : 2. Prove that one of the parallel sides is double the other.**

**Sol**. Since abcD is a trapezium,

âˆ´ AB y cÎ” â‡’ <1 = <2 and <3 = <4 (Alternate angle)

âˆ´ Î” apb ~ Î” cpD**Q2. In the figure, ABC is a D such that BD âŠ¥ AC and CE âŠ¥ AB. Prove that: PD Ã— BP = PC Ã— EP**

**Sol.** In Î” BEP and Î” CPD, we have:

<BPE = <CPD

[Vertically opp. angles]

<BEP = <CDP [Each = 90Â°]

âˆ´Using AA similarity, we have

Î” BEP ~ Î” CDP

âˆ´Their corresponding sides are proportional,

â‡’

â‡’ BP Ã— PD = EP Ã— PC

[By cross multiplication]**Q3. AB is a line segment. PB âŠ¥ AB and QA âŠ¥ AB are such that PO = 5 cm and QO = 8 cm. If ar (D POB) = 250 cm2, then find the area of D QOA.**

**Sol.** In Î” QOA and Î” POB,

<QOA = <BOP

[Vertically opposite angles]

<QAO = <PBO [Each = 90Â°]

âˆ´Using AA similarity, we have:**Q4. In the figure, ABCD is a trapezium in which AB y CD. If D BOC ~ D AOD, then prove that AD = BC.**

**Sol.** We have a trapezium ABCD in which AB y DC.

Since Î” BOC ~ Î” AOD [Given]

.(1)

In Î” ODC and Î”OBA,

<COD = <AOB

[Vertically opp. angles]

<ODC = <OBA [Alt. angles]

âˆ´ Using AA similarity, we have:

Î” ODC ~ Î” OBA

...(2)

From (1) and (2)

=

â‡’ OB Ã— OB = OA Ã— OA

â‡’ OB2 = OA2 â‡’ OA = OB ...(3)

From (1) to (3) we have**Q5. P and Q are points on sides of AB and AC respectively of Î” ABC. If AP = 3 cm, PB= 9 cm, AQ = 5 cm and QC = 15 cm, then show that BC = 4 PQ.**

**Sol**. We have Î” ABC in which P and Q are such that

AP = 3 cm, PB = 9 cm

AQ = 5 cm, QC = 15 cm

i.e., PQ divides AB and AC in the same ratio

âˆ´ PQ y BC

Now, in Î” APQ and Î” ABC

<P = <B [Corresponding angles]

<A = <A [Common]

â‡’ Using AA similarity,

Î” APQ ~ Î” ABC

â‡’ =

[Å’ AB = 3 + 9 = 12 cm and

AC = 5 + 15 = 20 cm]**Q6. On one of the longer sides PQ of a rectangle PQRS, a point O is taken such that SO ^{2} = POÂ· PQ Prove that: **Î”

Sol.

We have a rectangle PQRS such that

SO^{2} = POÂ· PQ

i.e., SO Ã— SO = PO Ã— PQ

â‡’ = ...(1)

[Å’ PQ = SR, opp. sides of rect. PQRS]

Now, in Î” POS and Î” OSR, we have:

<1 = <2 [OE PQ y SR,

opp. sides of a rect.]

â‡’ Using SAS similarity, we have

Î” POS ~ Î” OSR

**Q7. Determine the length of the altitude AD of an isosceles D ABC in which AB = AC = 2a cm and BC = a cm.**

**Sol.** We have D ABC in AD âŠ¥ BC and AB = AC = 2a. Also BC = a.

In D ADB and D ADC

<ADB = <ADC [Each = 90Â°]

<B = <C [Opp. angles to equal sides of a D]

âˆ´D ADB ~ D ADC

=

â‡’

Now in right Î” ABD, we have

AB^{2} = AD^{2} + BD^{2}

â‡’ AD^{2} = AB^{2} - BD^{2}

= (AB + BD) (AB - BD)

= **Q8. In an equilateral triangle with side â€˜aâ€™, prove that its area ****.**

**Sol. **We have Î” ABC such that

AB = BC = AC = a

Let us draw altitude AD âŠ¥ BC.

In Î” ADC and Î” ADB

AD = AD [Common]

AC = AB [Each = a]

<ADC = <ADB [Each = 90Â°]

âˆ´Î” ADC â‰Œ Î” ADB [RHS congruency]

âˆ´DC = DB =

Now, in right D ADB,

AB^{2} = AD^{2} + DB^{2}

â‡’ AD^{2} = AB^{2} - DB^{2}

= (AB + DB) (AB - DB)

=

Now, area of D ABC = 1/2Ã— Base Ã— altitude

Thus, the area of an equilateral triangle

= **Q9. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.****Sol.**We have Î” ABC in which AB = AC = CA and an altitude AD âŠ¥ BC.

In Î” ADB and Î” ADC

<ADB = <ADC [Each = 90Â°]

AB = AB [Given]

AD = AD [Common]

Using RHS congruency, we have

< ADB â‰Œ < ADC

â‡’ DB = DC = ...(1)

Now, in right Î” ADB, we have:

AB^{2} = AD^{2} + BD^{2} [Using Pythagoras Theorem]

â‡’ 3 [Side of the equilateral triangle]

= 4 [Altitude]2.**Q10. ABC is a right triangle in which <C = 90Â° and CD âŠ¥ AB. If BC = a, CA = b, AB = c and CD = p, then prove that: =**

**Sol. **We have a right Î” ABC such that <C = 90Â°.

Also, CD âŠ¥ AB

Now, ar (Î” ABC) =1/2 Ã— Base Ã— Height

= ...(1)

Also,

. .. (2)

From (1) and (2), we have

=

Dividing throughout by abp, we have:

=

Squaring both sides,

= = ...(3)

Now, In right D ABC,

AB^{2} = AC^{2} + BC^{2}

â‡’ c^{2} = b^{2} + a^{2} ...(4)

âˆ´ From (3) and (4), we get

=

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