Question

triangle abc ~ Triangle lmn in triangle abc ab= 5.1 cm angle b=40° bc= 4.8 cm ac/lm = 7/4 construct triangle abc and lmn​

Answer 1

Answer:

As shown in the figure,

Let B – C – N and B – A – L.

Rough Fighure

∆ABC ~ ∆LBN …[Given]

∴ ∠ABC ≅ ∠LBN …[Corresponding angles of similar triangles]

AB/LB = BC/CN = AC/LN …(i)[Corresponding sides of similar triangles]

But. = AC/LN = 4/7 …(ii)[Given]

∴ AB/LB = BC/CN = AC/LN = 4/7 …[From(i)and(ii)]

∴ sides of ∆LBN are longer than corresponding sides of ∆ABC.

∴ If seg BC is divided into 4 equal parts, then seg BN will be 7 times each part of seg BC.

So, if we construct ∆ABC, point N will be on side BC, at a distance equal to 7 parts from B.

Now, point L is the point of intersection of ray BA and a line through N, parallel to AC.

∆LBN is the required triangle similar to ∆ABC.

Steps of construction:

i. Draw ∆ABC of given measure. Draw ray BD making an acute angle with side BC.

ii. Taking convenient distance on compass, mark 7 points B1 , B2, B3 , B4 , B5 , B6 and B7 such that

BB1 = B1 B2 = B2 B3 = B3B4 = B4 B5 = B5 B6 = B6B7.

iii. Join B4 C. Draw line parallel to B7 C through B to intersects ray BC at N.

iv. Draw a line parallel to side AC through N. Name the point of intersection of this line and ray BA as L.

∆LBN is the required triangle similar to ∆ABC.

Answer 2

Answer:

9.00l of a gas at 16atm and 27 weigh 93.6 what is the molecular mass of the gas