Plane Geometry
- In ∆ PQR, L and M are two points on the sides PQ and PR respectively such that LM is parallel to QR. If PL = 2cm. LQ = 6 cm and PM = 1.5 cm, then MR (in cm) is
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According to question , we draw a figure triangle PQR
LM || QR
∴ ∠PLM = ∠PQR
∠PML = ∠PRQ
∴ By AA–similarity theorem,
∆PLM ~ ∆PQR∴ PL = PM ⇒ 2 = 1.5 LQ MR 6 MR
Correct Option: B
According to question , we draw a figure triangle PQR
LM || QR
∴ ∠PLM = ∠PQR
∠PML = ∠PRQ
∴ By AA–similarity theorem,
∆PLM ~ ∆PQR∴ PL = PM ⇒ 2 = 1.5 LQ MR 6 MR
⇒ 2MR = 1.5 × 6⇒ MR = 1.5 × 6 = 4.5 cm. 2
- In ∆ ABC, DE || AC, where D and E are two points lying on AB and BC respectively. If AB = 5 cm and AD = 3 cm, then BE : EC is
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We draw a figure triangle ABC in which DE || AC. D and E are two points lying on AB and BC respectively
In ∆ BDE and ∆ ABC,
DE || AC
∴ ∠BDE = ∠BAC
∠BED = ∠BCA
By AA–similarity,
∆BDE ~ ∆BAC⇒ DB = BE AD EC ⇒ AB - AD = BE AD EC ⇒ 5 - 3 = BE 3 EC
Correct Option: A
We draw a figure triangle ABC in which DE || AC. D and E are two points lying on AB and BC respectively
In ∆ BDE and ∆ ABC,
DE || AC
∴ ∠BDE = ∠BAC
∠BED = ∠BCA
By AA–similarity,
∆BDE ~ ∆BAC⇒ DB = BE AD EC ⇒ AB - AD = BE AD EC ⇒ 5 - 3 = BE 3 EC ⇒ BE = 2 EC 3
- In a ∆ ABC, if ∠A + ∠B = 135° and ∠C + 2 ∠B = 180°, then the correct relation is :
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On the basis of question we draw a figure of triangle ABC ,
Given that , ∠A + ∠B = 135°
We know that , ∠A + ∠B + ∠C = 180°
∴ ∠C = 180° – 135° = 45°
∴ ∠C + 2∠B = 180°
∴ 2∠B = 180° – ∠C = 180° – 45° = 135°∴ ∠B = 135° = 67.5° 2
Correct Option: A
On the basis of question we draw a figure of triangle ABC ,
Given that , ∠A + ∠B = 135°
We know that , ∠A + ∠B + ∠C = 180°
∴ ∠C = 180° – 135° = 45°
∴ ∠C + 2∠B = 180°
∴ 2∠B = 180° – ∠C = 180° – 45° = 135°∴ ∠B = 135° = 67.5° 2
∴ ∠A = ∠B
⇒ BC = AC
∠C < ∠B
⇒ AB < AC
- In ∆ ABC, ∠A = 90°, AD ⊥ BC and AD = BD = 2 cm. The length of CD is
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According to question , we draw a figure ∆ ABC
Given , ∠A = 90°, AD ⊥ BC and AD = BD = 2 cm.
In ∆ ABD,
∴ AB = √AD² + BD²
AB = √2² + 2²
AB = √8
AB = 2√2 cm.
∆ ABD ~ ∆BCA∴ AB = BD BC AB
⇒ AB² = BC × BD
⇒ (2√2)² = BC × 2
⇒ 8 = BC × 2
Correct Option: D
According to question , we draw a figure ∆ ABC
Given , ∠A = 90°, AD ⊥ BC and AD = BD = 2 cm.
In ∆ ABD,
∴ AB = √AD² + BD²
AB = √2² + 2²
AB = √8
AB = 2√2 cm.
∆ ABD ~ ∆BCA∴ AB = BD BC AB
⇒ AB² = BC × BD
⇒ (2√2)² = BC × 2
⇒ 8 = BC × 2⇒ BC = 8 = 4 cm. 2
∴ CD = BC – BD = (4 – 2) cm = 2 cm.
- B1 is a point on the side AC of ∆ABC and B1B is joined. A line is drawn through A parallel to B1B meeting BC at A1 and another line is drawn through C parallel to B1B meeting AB produced at C1. Then
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According to question , we draw a figure
In ∆ AA1C and ∆BB1C,
BB1 || AA1 ⇒ ∆AA1C ~ ∆BB1C∴ AA1 = AC ..... (i) BB1 B1C
In ∆ ACC1 and ∆ ABB1,
BB1 || CC1 ⇒ ∆ACC1 ~ ∆ABB1∴ CC1 = AC BB1 AB1 ⇒ BB1 = AB1 = AC - B1C CC1 AC AC ⇒ BB1 = 1 - B1C CC1 AC ⇒ BB1 = 1 - BB1 [From equation (i) CC1 AA1 ⇒ BB1 + BB1 = 1 CC1 AA1
Correct Option: B
According to question , we draw a figure
In ∆ AA1C and ∆BB1C,
BB1 || AA1 ⇒ ∆AA1C ~ ∆BB1C∴ AA1 = AC ..... (i) BB1 B1C
In ∆ ACC1 and ∆ ABB1,
BB1 || CC1 ⇒ ∆ACC1 ~ ∆ABB1∴ CC1 = AC BB1 AB1 ⇒ BB1 = AB1 = AC - B1C CC1 AC AC ⇒ BB1 = 1 - B1C CC1 AC ⇒ BB1 = 1 - BB1 [From equation (i) CC1 AA1 ⇒ BB1 + BB1 = 1 CC1 AA1 ⇒ 1 + 1 = 1 CC1 AA1 BB1
