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# Short Answer Type Questions: Triangles- 2 Class 10 Notes | EduRev

## Class 10 : Short Answer Type Questions: Triangles- 2 Class 10 Notes | EduRev

The document Short Answer Type Questions: Triangles- 2 Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
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Q11. ABC is a right triangle, right-angled at A, and D is the mid-point of AB. Prove that
BC2 = CD2 + 3 BD2 Sol. We have a right Δ ABC in which <A = 90°
∴Using Pythagoras Theorem, we have:
BC2 = AB2 + AC2   ...(1)
Again, Δ ACD is right D, <A = 90°
∴CD2 = AD2 + AC2
...(2) [Using Pythagoras Theorem]
Subtracting (2) from (1), we get
BC2 - CD2 = AB2 - AD2 ...(3)
Since D is the mid-point of AB
∴2 BD = AB  and  AD = BD ...(4)
From (3) and (4), we have:
∴BC2 - CD2 = (2 BD)2 - (BD)2
= 4 BD2 - BD2
BC2 = CD2 + 3 BD2

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:
CB2 = OF2 + BF2 ...(1)
In Δ OED
<E = 90°
∴Using Pythagoras theorem, we have:
OD2 = OE2 + DE2 ...(2)
Adding (1) and (2), we get
OB2 + OD2 = OF2 + BF2 + OE2 + DE2
= OF2 + AE2 + OE2 + CF2
[Œ BF = AE and CF = DE]
= (OF2 + CF2) + (OE2 + AE2)
= OC2 + OA2
= OC2 + 52
⇒ 62 + 82 = OC2 + 52
⇒ 36 + 64 = OC2 + 25
⇒ OC2 = 36 + 64 - 25 = 75 Thus OC = 5√3cm.

Q13. In the figure, if AD ⊥ Bc,  then prove that:
AB
2 + CD2 = AC2 + BD2 Sol. In D aDc, <ADc = 90°
∴ AD2 = AC2 - CD2 .....(1) (Using Pythagoras Theorem)
Similarly, in D AbD,
⇒ AD2 = AB2 - DB2.....(2)
From (1) and (2), we have
AB2 - DB2 = AC2 - CD2
⇒ AB2 + CD2 = AC2 + BD2

Q14. In the given figure, AD ⊥ BC and BD =  CD. Prove that:  Sol. BD = 1/2 CD
∴ 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
CA2 = AD2 + CD2  ...(1)
Also in right Δ ADB
AD2 = AB2 - BD2  ...(2)
From (1) and (2),
CA2 = AB2 - BD2+ CD2 Q15. In the given figure, M is the mid-point of side CD of parallelogram ABCD. The line BM is drawn intersecting AC at L, and AD produce D at E. Prove that EL = 2 BL. 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:
AB2 = BC2 + AC2
= 122 + 52
= 144 + 25 = 169
⇒AB = √169 = 13 cm
Now, from (1), we get ⇒ DE = = 2.77 cm
And = ⇒ AE Q17. In the given figure, DEFG is a square and <BAC = 90°. Show that DE2 = 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 × DG
But DG = EF = DE
∴BD × EC = DE × DE
⇒ BD × EC = DE2
Thus, DE2 = BD × EC

Q18. In the figure, AD ⊥ BC and BD =  CD. Prove that 2 CA2 = 2 AB2 + BC2 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,
CA2 = AD2 + DC2
= From (3) ...(4)
In right Δ ADB
Using Pythagoras theorem,
AD2 = AB2 - BD2
=   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 AB2 + AC2 = 2 [AD2 + BD2]. Sol. AD is a median,
∴  BD = DC
Let us draw AE ≌ BC
Now, in rt. Δs AEB and AEC, we have
AB2 = BE2 + AE2 ...(1)
AC2 = CE2 + AE2 ...(2)
Adding (1) and (2),
AB2 + AC2 = BE2 + AE2 + CE2 + AE2
= (BD - ED)2 + AE2 +
(CD + DE)2 + AE2
= 2AE2 + 2ED2 + BD2 + CD2
= 2 [AE2 + ED2] + BD2 + BD2
[BD = CD]
= 2 [AD]2 + 2BD2
[AE2 + ED2 = AD2]
= 2 [AD2 + BD2]
Thus,  AB2 + AC2 = 2 [AD2 + BD2]

Q23. Triangle ABC is right angled at B and D is the mid point of BC. Prove that:
AC2 = 4 AD- 3 AB2
Sol.
D is the mid-point of BC.
∴ BC = 2 BD Now, in Δ ABC
AC2 = BC2 + AB2
= (2 BD)2 + AB2
= 4 BD2 + AB2 ...(1)
In right Δ ABD
Using Pythagoras theorem,
AD2 = AB2 + BD2
⇒ BD2 = AD2 - AB2 ...(2)
From (1) and (2), we get
AC2 = 4 [AD2 - AB2] + AB2
⇒ AC2 = - 4 AB2 + 4 AD2 + AB2
⇒ AC2 = - 3 AB2 + 4 AD2
or AC2 = 4 AD2 - 3 AB2

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