1. A ABC in which AB
2. A PQR in which PQ=5.5 cm, QR = 6.5 cm, RP = 5 cm.
3. A XYZ in which XZ = 8.4 cm, XY = 6.8 cm, YZ = 7.5 cm.
4. A DEF in which DE = 8 cm, DF = 7.2 cm, EF = 6.3 cm.
5. A LMN in which LN = 7 cm, NM=5.5 cm, LM= 6.4 cm
Answers
1 . ABC is a triangle in which AB = Ac and D is a point on the side AC such that BC2 = AC × CD. Prove that BD = BC. ABC is a triangle in which AB = Ac and D is a point on the side AC such that BC2 = AC × CD
2. taking P as centre and measure 5cm , cut an arc . taking Q as centre and measure 6.5 cm , cut an arc name the point meeting as R . PQR is the required triangle .
3. Draw a line segment which is sufficiently long using ruler.
(ii) Locate points X and Z on it such that XZ=9.5cm.
(iii) With X as centre and radius 7.8cm, draw an arc(see figure)
(iv) With Z as centre and radius 4.5cm , draw another arc cutting the previous arc at C.
(v) Join XY and YZ
4
Step 1: Draw DE=8cm
Step 2: With D as centre and radius =7.2cm, draw an arc at F.
Step 3: With E as centre, and radius =6.3cm, draw an arc at F.
Step 4: Join DF and EF.
Hence, ΔDEF is the required triangle.
5 . Construct ΔLMN in which LN=7 cm, NM=5.5 cm, LM=6.4 cm.
6 0.04 metre
Step-by-step explanation:
Answer:
\large{\underline{\underline{{\sf{\purple{Given \:Equations:-}}}}}}
GivenEquations:−
➨ 2x + 3y = -5
➨ x + y = -1
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Putting the value of x = 0 in equation 2x + 3y = -5 :-
➨ 2(0) + 3y + 5 = 0
➨ y = \dfrac{-5}{3}
3
−5
⚘ The Coordinates are ( 0 , \frac{-5}{3}
3
−5
)
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Putting value of y = 0 in equation 2x + 3y = -5 :-
➨ 2x + 3(0) + 5 = 0
➨ x = \dfrac{-5}{2}
2
−5
⚘ Their Coordinates are ( \frac{-5}{2}
2
−5
,0)
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Putting the value of x = 0 in equation x + y = -1 :-
➨ 0 + y + 1 = 0
➨ y = -1
⚘ Their Coordinates are ( 0 , -1 )
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Putting the value of y = 0 in equation x + y = -1 :-
➨ x + 0 + 1 = 0
➨ x = -1
⚘ Their Coordinates are ( -1 , 0 )
Both lines intersect at a point (2 , -3).
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