Question

Question:

Construct APQR if PQ = 6.5 cm, mzPQR = 110° and m<QRP = 40°. (Hint: Recall angle sum property of a triangle).​

Answer 1

$\begin{gathered}{\textsf{\textbf{\underline{\underline{ Answer :}}}}}\end{gathered}$

⟹ We have to construct a triangle.

$\begin{gathered}{\textsf{\textbf{\underline{\underline{ Given :}}}}}\end{gathered}$

PQ = 6.5cm

<PQR = 110°

<QRP = 40°

⟹ First we have to find <P

$\begin{gathered}{\textsf{\textbf{\underline{\underline{ ⟹We know that :}}}}}\end{gathered}$

⟹ The sum of angles of triangle is 180°

<PQR + <QRP +<RPQ = 180°

110° + 40° + <RPQ = 180°

150° + <RPQ =180°

<RPQ = 180°-150°

<RPQ = 30°

⟹ NOW, we have to construct a ∆PQR.

$\begin{gathered}{\textsf{\textbf{\underline{\underline{Note :}}}}}\end{gathered}$

⟹ See the construction of traingle ∆PQR in above attachment.

⟹ First we make rough sketch of ∆PQR.

✮:▹Steps of construction:

⟹ 1. Draw a line segment PQ = 6.5cm

⟹ 2. With P as a centre draw an angle of 30°

⟹ 3. With Q As a centre draw an angle of 110° to meet each other at R

∴the required ∆PQR

_______________________________

Answer 2

Answer:

Solution:

We use the basic rules of construction to solve the question given.

Let's use the angle-sum property of a triangle to find the measure of ∠RPQ in ∆PQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°

Given that, m∠PQR = 105° and m∠QRP = 40°

∠PQR + ∠QRP + ∠RPQ = 180°.

105° + 40° + ∠RPQ = 180°

So, ∠RPQ = 35°

Now, let’s construct ΔPQR such that PQ = 5cm, ∠PQR = 105° and ∠RPQ = 35°, with the steps given below

Construct APQR if PQ = 6.5 cm, mzPQR = 110° and m<QRP = 40°. (Hint: Recall angle sum property of a triangle).Steps of construction :

Draw a line segment PQ of length 5 cm.

At P, draw a ray PX making 35° with PQ.

At Q, draw a ray QY making 105° with PQ.

Rays PX and QY will intersect at point R.

Triangle PQR is now constructed

Step-by-step explanation:

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