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

Class 10 Mathematics by VP Classes

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.
All you need of Class 10 at this link: Class 10

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

Sol. Since abcD is a trapezium, 
∴ AB y cΔ ⇒ <1 = <2 and <3 = <4 (Alternate angle)
∴  Δ apb ~ Δ cpD
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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

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

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,
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
⇒ 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.
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Sol. In Δ QOA and Δ POB,
<QOA = <BOP
[Vertically opposite angles]
<QAO = <PBO [Each = 90°]
∴Using AA similarity, we have:
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Q4. In the figure, ABCD is a trapezium in which AB y CD. If D BOC ~ D AOD, then prove that AD = BC.
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Sol. We have a trapezium ABCD in which AB y DC.
Since Δ BOC   ~ Δ AOD [Given]
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev .(1)
In Δ ODC and ΔOBA,
<COD = <AOB
[Vertically opp. angles]
  <ODC = <OBA [Alt. angles]
∴ Using AA similarity, we have:
Δ ODC ~ Δ OBA
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev  ...(2)
From (1) and (2)
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
⇒ OB × OB = OA × OA
⇒ OB2 = OA2  ⇒  OA = OB  ...(3)
From (1) to (3) we have
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
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
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
⇒ =
[ΠAB = 3 + 9 = 12 cm and
AC = 5 + 15 = 20 cm]
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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.

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

We have a rectangle PQRS such that
SO2 = PO· PQ
i.e., SO × SO = PO × PQ
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
⇒ =  ...(1) 
[ΠPQ = SR, opp. sides of rect. PQRS]
Now, in Δ POS and Δ OSR, we have:
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
<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.
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Sol. We have D ABC in AD ⊥ BC and AB = AC = 2a. Also BC = a.
In D ADB and D ADC
<ADB = <ADC   [Each = 90°]
<B = <C [Opp. angles to        equal sides of a D]
∴D ADB ~ D ADC
=   Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
⇒   Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev  
Now in right Δ ABD, we have
AB2 = AD2 + BD2
⇒ AD2 = AB2 - BD2
= (AB + BD) (AB - BD)
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

Q8. In an equilateral triangle with side ‘a’, prove that its area  Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev.
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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 D ADB,
AB2 = AD2 + DB2
⇒ AD2 = AB2 - DB2
= (AB + DB) (AB - DB)
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Now, area of D ABC =  1/2× Base × altitude
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Thus, the area of an equilateral triangle
=  Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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.
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
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 = Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev  ...(1)
Now, in right Δ ADB, we have:
AB2 = AD2 + BD2 [Using Pythagoras Theorem]
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
⇒ 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:
 =
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev 

Sol. We have a right Δ ABC such that <C = 90°.
Also, CD ⊥ AB
Now, ar (Δ ABC) =1/2 × Base × Height
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev   ...(1)
Also, Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev   
Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev . .. (2)
From (1) and (2), we have
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Dividing throughout by abp, we have:
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev
Squaring both sides,
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev  =       ...(3)
Now, In right D ABC,
AB2 = AC2 + BC2 
⇒ c2 = b2 + a2         ...(4)
∴ From (3) and (4), we get
= Short Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRevShort Answer Type Questions (Part 1) - Triangles Class 10 Notes | EduRev

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