The document Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.

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

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

∴ AB y cΔ ⇒ <1 = <2 and <3 = <4 (Alternate angle)

∴ Δ APB ~ Δ CPD**Q2. In the figure, ABC is a D such that BD ⊥ AC and CE ⊥ AB. Prove that: PD × BP = PC × EP**

**Sol.** In Δ BEP and Δ CPD, we have:

<BPE = <CPD

[Vertically opp. angles]

<BEP = <CDP [Each = 90°]

∴Using AA similarity, we have

Δ BEP ~ Δ CDP

∴Their corresponding sides are proportional,

⇒

⇒ BP × PD = EP × PC

[By cross multiplication]**Q3. AB is a line segment. PB ⊥ AB and QA ⊥ 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 Δ QOA and Δ POB,

<QOA = <BOP

[Vertically opposite angles]

<QAO = <PBO [Each = 90°]

∴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 Δ BOC ~ Δ AOD [Given]

.(1)

In Δ ODC and ΔOBA,

<COD = <AOB

[Vertically opp. angles]

<ODC = <OBA [Alt. angles]

∴ Using AA similarity, we have:

Δ ODC ~ Δ OBA

...(2)

From (1) and (2)

=

⇒ OB × OB = OA × OA

⇒ OB2 = OA2 ⇒ OA = OB ...(3)

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

**Sol**. We have Δ 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

∴ PQ y BC

Now, in Δ APQ and Δ ABC

<P = <B [Corresponding angles]

<A = <A [Common]

⇒ Using AA similarity,

Δ APQ ~ Δ ABC

⇒ =

[Œ 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} = PO· PQ Prove that: **Δ

Sol.

We have a rectangle PQRS such that

SO^{2} = PO· PQ

i.e., SO × SO = PO × PQ

⇒ = ...(1)

[Œ PQ = SR, opp. sides of rectangle PQRS]

Now, in Δ POS and Δ OSR, we have:

<1 = <2 [OE PQ y SR,

opp. sides of a rectangle]

⇒ Using SAS similarity, we have

Δ POS ~ Δ 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 Δ ABC in AD ⊥ BC and AB = AC = 2a. Also BC = a.

In Δ ADB and Δ ADC

<ADB = <ADC [Each = 90°]

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

∴Δ ADB ~ Δ ADC

=

⇒

Now in right Δ ABD, we have

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

⇒ AD^{2} = AB^{2} - BD^{2}

= (AB + BD) (AB - BD)

= **Q8. In an equilateral triangle with side ‘a’, prove that its area ****.**

**Sol. **We have Δ ABC such that

AB = BC = AC = a

Let us draw altitude AD ⊥ BC.

In Δ ADC and Δ ADB

AD = AD [Common]

AC = AB [Each = a]

<ADC = <ADB [Each = 90°]

∴Δ ADC ≌ Δ ADB [RHS congruency]

∴DC = DB =

Now, in right Δ ADB,

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

⇒ AD^{2} = AB^{2} - DB^{2}

= (AB + DB) (AB - DB)

=

Now, area of Δ ABC = 1/2× Base × 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 Δ ABC in which AB = AC = CA and an altitude AD ⊥ BC.

In Δ ADB and Δ ADC

<ADB = <ADC [Each = 90°]

AB = AB [Given]

AD = AD [Common]

Using RHS congruency, we have

< ADB ≌ < ADC

⇒ DB = DC = ...(1)

Now, in right Δ ADB, we have:

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

⇒ 3 [Side of the equilateral triangle]

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

**Sol. **We have a right Δ ABC such that <C = 90°.

Also, CD ⊥ AB

Now, ar (Δ ABC) =1/2 × Base × 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}

⇒ c^{2} = b^{2} + a^{2} ...(4)

∴ From (3) and (4), we get

=

**Q11. ABC is a right triangle, right-angled at A, and D is the mid-point of AB. Prove that BC ^{2} = CD^{2} + 3 BD^{2}. **

**Sol. **We have a right Δ ABC in which <A = 90°

∴Using Pythagoras Theorem, we have:

BC^{2} = AB^{2} + AC^{2} ...(1)

Again, Δ ACD is right D, <A = 90°

∴CD^{2} = AD^{2} + AC^{2}

...(2) [Using Pythagoras Theorem]

Subtracting (2) from (1), we get

BC^{2} - CD^{2} = AB^{2} - AD^{2} ...(3)

Since D is the mid-point of AB

∴2 BD = AB and AD = BD ...(4)

From (3) and (4), we have:

∴BC^{2} - CD^{2} = (2 BD)^{2} - (BD)^{2}

= 4 BD^{2} - BD^{2}

BC^{2} = CD^{2} + 3 BD^{2}**Q12. In the figure, O is any point inside a rectangle ABCD such that OB = 6 cm, OD = 8 cm and OA = 5 cm. Find the length of OC. **

**Sol.** Let us draw EOF y AB ⇒ OE ⊥ AD and OF ⊥ BC

In Δ OFB

Œ <F = 90°

∴Using Pythagoras theorem, we have:

CB^{2} = OF^{2} + BF^{2} ...(1)

In Δ OED

<E = 90°

∴Using Pythagoras theorem, we have:

OD^{2} = OE^{2} + DE^{2} ...(2)

Adding (1) and (2), we get

OB^{2} + OD^{2} = OF^{2} + BF^{2} + OE^{2} + DE^{2}

= OF^{2} + AE^{2} + OE^{2} + CF^{2}

[Œ BF = AE and CF = DE]

= (OF^{2} + CF^{2}) + (OE^{2} + AE^{2})

= OC^{2} + OA^{2}

= OC^{2} + 52

⇒ 62 + 82 = OC^{2} + 52

⇒ 36 + 64 = OC^{2} + 25

⇒ OC^{2} = 36 + 64 - 25 = 75

⇒ Thus OC = 5√3cm.**Q13. In the figure, if AD ⊥ Bc, then prove that: AB**

**Sol.** In D aDc, <ADc = 90°

∴ AD^{2} = AC^{2} - CD^{2} .....(1) (Using Pythagoras Theorem)

Similarly, in D AbD,

⇒ AD^{2} = AB^{2} - DB^{2}.....(2)

From (1) and (2), we have

AB^{2} - DB^{2} = AC^{2} - CD^{2}

⇒ AB^{2} + CD^{2} = AC^{2} + BD^{2}**Q14. In the given figure, AD ⊥ BC and BD = CD. Prove that:**

∴ 3 BD = CD

Since BD + DC = BC

∴ BD + 3 BD = BC

⇒ 4 BD = BC

⇒ BD = 1/4 BC

⇒ CD = 3/4 BC

Now, in right Δ ADC, < D = 90°

By Pythagoras theorem, we get

CA

Also in right Δ ADB

AD

From (1) and (2),

CA

**Sol. **We have parallelogram ABCD in which M is the mid point of CD.

In Δ EMD and Δ BMC

MD = MC [Œ M is mid-point of CD]

<EMD = <CMB

[Vertically opposite angles]

<MED = <MBC

[Alternate interior angles]

∴ Δ BMC ≌ Δ EMD [AAS congruency]

⇒ BC = ED ⇒ AD = ED ...(1)

[Œ BC = AD, opposite sides of parallelogram]

Now, in Δ AEL and Δ CBL

<AEL = <CBL [Alternate interior angles]

<ALE = <CLB

[Vertically opposite angles]

∴ By AA similarity, we have:**Q16. In the given figure, Δ ABC is right-angled at C and DE ^ AB. Prove that Δ ABC ~ Δ ADE and hence find the length of AE and DE.**

Sol.In Δ ABC and Δ ADE, we have:<A = <A [Common]

<C = <E [Each = 90°]

∴ Δ ABC ~ Δ ADE [AA Similarity]

...(1)In right Δ ABC, <C = 90°Using Pythagoras theorem, we have:

AB

= 122 + 52

= 144 + 25 = 169

⇒AB = √169 = 13 cm

Now, from (1), we get

⇒ DE = = 2.77 cmAnd = ⇒ AE

**Q17. In the given figure, DEFG is a square and <BAC = 90°. Show that DE ^{2} = BD × EC.**

**Sol.** In Δ DBG and Δ ECF

<3 + <1 = 90° = <3 + <4

∴<3 + <1 = <3 + <4

⇒ <1 = <4

<D = E = 90°

∴Using AA similarity, we have:

= BD × EC = EF × DGBut DG = EF = DE

∴BD × EC = DE × DE

⇒ BD × EC = DE^{2}

Thus, DE^{2} = BD × EC**Q18. In the figure, AD ⊥ BC and BD = CD. Prove that 2 CA ^{2} = 2 AB^{2} + BC^{2}. **

**Sol.** Œ BD = 1/3 CD

⇒ 3 BD = CD

∴BC = BD + DC

⇒ BC = BD + 3 BD

⇒ BC = 4 BD ...(1)

And From (2) ...(4)In right Δ ADC,

Using Pythagoras theorem,

CA^{2} = AD^{2} + DC^{2}

= From (3) ...(4)In right Δ ADB

Using Pythagoras theorem,

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

= **Q19.If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.**

**Sol.** We have a quadrilateral ABCD such that its diagonals intersect at O and

= <AOB = <COD [Vertically opposite angles]∴ Using SAS similarity, we have

Δ AOB ~ Δ COD

⇒ Their corresponding angles are equal.

i.e., <1 = <2

But they form a pair of int. alt. angles.

⇒ AB y DC

⇒ ABCD is a trapezium.**Q20. Two triangles ABC and DBC are on the same base BC and on the same side of BC in which <A = <D = 90°. If CA and BD meet each other at E, show that AE· EC = BE· ED Sol.** We have right Δ ABC and right Δ DBC on the same base BC such that

<A = <D = 90°

In Δ ABE and Δ DCE

<A = <D = 90°

<1 = <2 [Vertically opp. angles]

∴ Using AA similarity, we have:

Δ ABE ~ Δ DCE

⇒ Their corresponding sides are proportional.

⇒ ⇒ AE· EC = BE· ED**Q21. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Δ ABE ~ Δ CFB Sol. **We have parallelogram ABCD

In Δ ABE and Δ CFB, we have

<A = <C

[Opposite angles of parallelogram]

<AEB = <EBC

[Alternate angles, AD y BC]

∴ Using AA similarity, we get

Δ ABE ~ Δ CFB

**Q22. In Δ ABC, if AD is the median, then show that AB ^{2} + AC^{2} = 2 [AD^{2} + BD^{2}]. **

∴ BD = DC

Let us draw AE ≌ BC

Now, in rt. Δs AEB and AEC, we have

AB

AC

Adding (1) and (2),

AB

= (BD - ED)2 + AE2 +

(CD + DE)2 + AE2

= 2AE

= 2 [AE

[BD = CD]

= 2 [AD]2 + 2BD

[AE

= 2 [AD

Thus, AB

AC

Sol.

∴ BC = 2 BD

Now, in Δ ABCAC

= (2 BD)

= 4 BD

In right Δ ABD

Using Pythagoras theorem,

AD

⇒ BD

From (1) and (2), we get

AC

⇒ AC

⇒ AC

or AC

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