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# Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

## Class 10 : Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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 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
SO2 = PO· PQ
Prove that:
Δ POS ~ Δ OSR.
Sol. We have a rectangle PQRS such that
SO2 = PO· PQ
i.e., SO × SO = PO × PQ ⇒ =  ...(1)
[Œ PQ = SR, opp. sides of rect. PQRS]
Now, in Δ POS and Δ OSR, we have: <1 = <2 [OE PQ y SR,
opp. sides of a rect.]
⇒ 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 D ABC in AD ⊥ BC and AB = AC = 2a. Also BC = a.
<B = <C [Opp. angles to        equal sides of a D]
= ⇒ Now in right Δ ABD, we have
⇒ AD2 = AB2 - BD2
= (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.
AC = AB [Each = a]
∴DC = DB =
⇒ AD2 = AB2 - DB2
= (AB + DB) (AB - DB)
= Now, area of D 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.
AB = AB [Given]
Using RHS congruency, we have
⇒ DB = DC = ...(1)
Now, in right Δ ADB, we have:
AB2 = AD2 + BD2 [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,
AB2 = AC2 + BC2
⇒ c2 = b2 + a2         ...(4)
∴ From (3) and (4), we get
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