Question #59194

2 Answers
Feb 4, 2018

As explainedenter image source here

Explanation:

Since triangles ABC & AQR are similar,

(AQ )/ (AB) = (AR) / (AC) = (QR) / (BC)AQAB=ARAC=QRBC

Also AhatBC = AhatQR = 60^0AˆBC=AˆQR=600

But AQ = AB * 1.5 = 5 * 1.5 = 7.5 cm#

From the diagram,
BC = (CD) / sin 60 = 3 / (sqrt3/2) = 2 sqrt3 = 3.4641#

QR = (BC) * 1.5 = 3.4641 * 1.5 = 5.196 cm#

Now we know two sides AQ, QR and included angle hatQˆQ

Now we can construct Delta AQR

Feb 4, 2018

drawndrawn

Steps for construction

  1. Line segments of length 5cm and 3cm are drawn with the help of a ruler.
  2. A line segment AX of length > 8cm is also drawn similarly.
  3. Line segment AB is then cut off from AX with the help of a compass.
  4. /_ABK=60^@ is drawn with the help of a ruler and a compass.
  5. With the help of a ruler and a compass a perpendicular AN on AB at A is drawn and an intercept AP=3cm is cut off with the help of a compass.
  6. PO is drawn parallel to AB with the help of a compass and a ruler. PO intersects BK at C. A and C are joined . The DeltaABC is the required triangle having altitude CD=3cm.
  7. AB is bisected at E with the help of a compass and a ruler.
  8. Line segment BQ=BE is cut off from extended part of AB to make AQ=1.5xxAB.
  9. /_AQL=60^@ is drawn to get get QL"||"BC. This QL intersects extended AC at R.
  10. Now DeltaAQR is the similar to DeltaABC having 1.5 times of the corresponding sides of DeltaABC.